CHAPTER 4 The Laplace Transform 4.1 Introduction The Laplace transform provides an effective method of solving initial-value problems for linear differential equations with constant coefficients. However, the usefulness of Laplace transforms is by no means restricted to this class of problems. Some understanding of the basic theory is an essential part of the mathematical background of engineers, scientists and mathematicians. The Laplace transform is defined in terms of an integral over the interval [0, ∞). In- tegrals over an infinite interval are called improper integrals, a topic studied in Calculus II. DEFINITION Let f be a continuous function on [0, ∞). The Laplace transform of f, denoted by L[(f(x)], or by F (s), is the function given by ∞ L[f(x)] = F (s)=Z e−sxf(x) dx. (1) 0 The domain of F is the set of all real numbers s for which the improper integral converges. In more advanced treatments of the Laplace transform the parameter s assumes com- plex values, but the restriction to real values is sufficient for our purposes here. Note that L transforms a function f = f(x) into a function F = F (s) of the parameter s. The continuity assumption on f will hold throughout the first three sections. It is made for convenience in presenting the basic properties of L and for applying the Laplace transform method to solving initial-value problems. In the last two sections of this chapter we extend the definition of L to a larger class of functions, the piecewise continuous functions on [0, ∞). There we will apply L to the problem of solving nonhomogeneous equations in which the nonhomogeneous term is piecewise continuous. This will involve some extension of our concepts of differential equation and solution. As indicated above, the primary application of Laplace transforms of interest to us is solving linear differential equations with constant coefficients. Referring to our work in Chapter 3, the functions which arise naturally in the treatment of these equations are: p(x)erx,p(x) cos βx, p(x) sin βx, p(x)erx cos βx, p(x)erx sin βx where p is a polynomial. We begin by calculating the Laplace transforms of some simple cases of these functions. 115 Example 1. Let f(x)=1· e0x ≡ 1on[0, ∞). By the Definition, ∞ b L[1] = Z e−sx · 1 dx = lim Z e−sx dx 0 b→∞ 0 − b − " e sx # e sb 1 −1 1 = lim = lim + = lim + . b→∞ −s b→∞ −s s b→∞ sesb s 0 Now, lim −1/sesb exists if and only if s>0, and in this case b→∞ −1 lim =0. b→∞ sesb Thus, 1 L[1] = ,s>0. s Example 2. Let f(x)=erx on [0, ∞). Then, ∞ b L[erx]=Z e−sx · erx dx = lim Z e−(s−r)x dx 0 b→∞ 0 b −(s−r)x " −(s−r)b # e e 1 = lim = lim + . b→∞ −(s − r) b→∞ −(s − r) s − r 0 The limit exists (and has the value 0) if and only if s − r>0. Therefore 1 L[erx]= , s>r. s − r Note that if r = 0, then we have the result in Example 1. Example 3. Let f(x) = cos βx on [0, ∞). Then, ∞ b L[cos βx]=Z e−sx · cos βx dx = lim Z e−sx cos βx dx 0 b→∞ 0 − b e sx[−s cos βx − β sin βx] = lim . b→∞ s2 + β2 0 (Note the integral was calculated using integration by parts; also, it is a standard entry in a table of integrals.) Now, 1 s cos βb + β sin βb s L[cos βx]=− lim · + . b→∞ e−sb s2 + β2 s2 + β2 Since [s cos βb+ β sin βb]/(s2 + β2) is bounded, the limit exists (and has the value 0) if and only if s>0. Therefore, s L[cos βx]= ,s>0. s2 + β2 116 The following table gives a basic list of the Laplace transforms of functions that we will encounter in this chapter. While the entries in the table can be verified using the Definition, some of the integrations involved are complicated. The properties of the Laplace transform presented in the next section provide a more efficient way to obtain many of the entries in the table. Handbooks of mathematical functions, for example the CRC Standard Mathematical Tables, give extensive tables of Laplace transforms. Table of Laplace Transforms f(x) F (s)=L[f(x)] 1 1 ,s>0 s 1 eαx , s>α s − α s cos βx ,s>0 s2 + β2 β sin βx ,s>0 s2 + β2 s − α eαx cos βx , s>α (s − α)2 + β2 β eαx sin βx , s>α (s − α)2 + β2 n! xn,n=1, 2,... ,s>0 sn+1 n! xn erx,n=1, 2,... , s>r (s − r)n+1 s2 − β2 x cos βx ,s>0 (s2 + β2)2 2βs x sin βx ,s>0 (s2 + β2)2 Exercises 4.1 Use the definition of the Laplace transform to find the Laplace transform of the given function. 1. f(x)=x. 2. f(x)=x2. 117 3. f(x) = sin x. 4. f(x)=xerx. 5. f(x) = sinh x. 6. f(x) = cosh x. eax 7. Use the fact that Z eax cos bx dx = [a cos bx + b sin bx] to find L[erx cos βx] a2 + b2 by the definition. eax 8. Use the fact that Z eax sin bx dx = [a sin bx − b cos bx] to find L[erx sin βx] a2 + b2 by the definition. 2π Z e−sx sin xdx 9. Show that L[sin x]= 0 . 1 − e−2πs 10. Let f be a continuous function on [0, ∞) and suppose that f is periodic with period p. That is f is continuous and f(x + p)=f(x),p>0, for all x. Show that p Z e−sxf(x) dx L[f(x)] = 0 . 1 − e−ps 118 4.2 Basic Properties of the Laplace Transform In the preceding section we defined the Laplace transform and calculated the Laplace trans- forms of some of the functions that occur in solving linear differential equations with con- stant coefficients. In this section we consider the basic question of the existence of the Laplace transform of a function f, and we develop the properties of the Laplace transform that will be used in solving initial value problems. To motivate the material in this section, consider the differential equation y00 + ay0 + by = f(x) (2) where a and b are constants and f is a continuous function on [0, ∞). If we assume that y = y(x) is a solution of (1) and formally apply L, we obtain 00 0 L y (x)+ay (x)+by(x) = L[f(x)]. (3) The right-hand side of this equation suggests the basic question of the existence of L[f(x)]. That is, for what functions f does L[f] exist? DEFINITION A function f, continuous on [0, ∞), is said to be of exponential order λ, λ a real number, if there exists a positive number M and a nonnegative number A such that | f(x)|≤Meλx on [A, ∞). Example 1. (a) If f is a bounded function on [0, ∞) [for example, f(x) = cos βx or f(x)=sinβx], then f is of exponential order 0. f bounded implies that there exists a positive number M such that | f(x)|≤M for all x ∈ [0, ∞). Therefore, | f(x)|≤M = Me0x on [0, ∞). [Note: if f(x) = cos βx or f(x) = sin βx, then we could take M = 1.] (b) Let f(x)=x on [0, ∞). For any positive number λ, x lim =0 x→∞ eλx by L’Hˆopital’srule. Therefore, there exists a nonnegative number A such that x ≤ 1on[A, ∞). eλx This implies that x ≤ eλx =1· eλx on [A, ∞) and f(x)=x is of exponential order λ. The same argument can be used to show that f(x)=xα,α any real number, is of exponential order λ for any positive number λ. In general, if p = p(x) is a polynomial, then p is of exponential order λ for any positive number λ. 119 (c) If f(x)=erx, then f is of exponential order λ for any λ ≥ r. (d) Consider the function f(x)=ex2 .If f is of exponential order λ for some λ, then there exists a positive number M and a nonnegative number A such that 2 2 ex ≤ Meλx on [A, ∞) which implies e−λxex ≤ M on [A, ∞). But, 2 lim e−λxex = lim ex(x−λ) = ∞, x→∞ x→∞ a contradiction. Thus f(x)=ex2 is not of exponential order λ for any positive number λ. Our first property of L is a sufficient condition for L[f(x)] to exist. We shall omit the proof. THEOREM 1. Let f be a continuous function on [0, ∞). If f is of exponential order λ, then the Laplace transform L[f(x)] = F (s) exists for s>λ. We now turn to the left-hand side of equation (2) where we have L applied to the linear combination y00(x)+ay0(x)+by(x). THEOREM 2. The operator L is a linear operator. That is, if g and h are continuous functions on [0, ∞), and if each of L[g(x)] and L[h(x)] exists for s>λ, then L[g(x)+h(x)] and L[cg(x)],c constant, each exist for s>λ, and L[g(x)+h(x)] = L[g(x)] + L[h(x)] L[cg(x)] = cL[g(x)].